Problem: Divide the following complex numbers. $\dfrac{11+24i}{-4-i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-4+i}$. $ \dfrac{11+24i}{-4-i} = \dfrac{11+24i}{-4-i} \cdot \dfrac{{-4+i}}{{-4+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(11+24i) \cdot (-4+i)} {(-4)^2 - (-i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(11+24i) \cdot (-4+i)} {(-4)^2 - (-i)^2} $ $ = \dfrac{(11+24i) \cdot (-4+i)} {16 + 1} $ $ = \dfrac{(11+24i) \cdot (-4+i)} {17} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({11+24i}) \cdot ({-4+i})} {17} $ $ = \dfrac{{11} \cdot {(-4)} + {24} \cdot {(-4) i} + {11} \cdot {1 i} + {24} \cdot {1 i^2}} {17} $ $ = \dfrac{-44 - 96i + 11i + 24 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{-44 - 96i + 11i - 24} {17} = \dfrac{-68 - 85i} {17} = -4-5i $